Discussion
Consider two infinitely long thin concentric tubes of circular cross section as shown in figre. If D1 and D2 are the diameters of the inner and outer tubes respectively, then the view factor F22 is given by:
Answer: D
The view factor is \(F_{22} = 1 - \left(\frac{D_{1}}{D_{2}}\right)\).
All radiation leaving the inner tube reaches the outer tube, so \(F_{12} = 1\). By reciprocity, \(A_{2}F_{21} = A_{1}F_{12}\), and per unit length the areas are in the ratio of the diameters, so \(F_{21} = D_{1}/D_{2}\). Since the outer tube's inside surface is concave, it also sees itself, and the factors from surface 2 must add up to one:
\[ F_{22} = 1 - F_{21} = 1 - \frac{D_{1}}{D_{2}} \]
\(D_{1}/D_{2}\) alone is \(F_{21}\), not \(F_{22}\). Zero would be right only for a flat or convex surface, and \(D_{2}/D_{1} - 1\) can exceed one.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development